Question: Let $R$ be the region enclosed by the line $y=-1$, the line $x=2$, and the curve $y=x^2-1$. $y$ $x$ ${y=x^2-1}$ $ R$ $ 0$ $ 2$ $ -1$ A solid is generated by rotating $R$ about the line $y=-1$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_{-1}^3 (x^2-1)^2\, dx$ (Choice B) B $\pi \int_0^2 (x^2-1)^2\, dx$ (Choice C) C $\pi \int_{-1}^3 x^4\, dx$ (Choice D) D $\pi \int_0^2 x^4\, dx$
Explanation: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=x^2-1}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=x^2-1}$ $ 0$ $ 2$ $ -1$ $r$ The radius is equal to the distance between the curve $y=x^2-1$ and the line $y=-1$. In other words, for any $x$ -value, this is the equation for $r(x)$ : $\begin{aligned} r(x)}&=(x^2-1)-(-1) \\\\ &=x^2} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(x^2}\right)^2 \\\\ &=\pi x^4 \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=2$. So the interval of integration is $\left[0,2\right]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int^{2}_0 \left(\pi x^4\right)\, dx \\\\ &=\pi\int^{2}_0 x^4\, dx \end{aligned}$